Question

In a screw gauge the zero of main scale reference line coincides with the fifth division of the circular scale when two studs are in contact. There are 100 divisions in circular scale and pitch of screw gauge is 0.1 mm. When diameter of a sphere is measured, the reading of main scale is 5 mm and 50th division of circular scale coincides with the reference line of main scale. The diameter of sphere is _______ mm.}

• A. 5.045
• B. 5.055
• C. 5.450
• D. 5.550

Answer 1

The Correct Option is B

The screw gauge gives the reading in the following way: - Main scale reading (MSR) = 5 mm, - Circular scale reading (CSR) = 50 divisions. Now, the pitch of the screw gauge is 0.1 mm, and the total number of divisions in the circular scale is 100. So, the least count (LC) of the screw gauge is: \[ LC = \frac{\text{Pitch of screw gauge}}{\text{Number of divisions in circular scale}} = \frac{0.1}{100} = 0.001 \, \text{mm}. \] The total reading is given by: \[ \text{Diameter of sphere} = \text{MSR} + (\text{CSR} \times LC). \] Substitute the known values: \[ \text{Diameter of sphere} = 5 + (50 \times 0.001) = 5 + 0.05 = 5.055 \, \text{mm}. \]
Final Answer: 5.055 mm